Questions

The perpendicular form of the line $\sqrt{3} x - y + 4 = 0$ is

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a

x \cos \frac{5 π}{6} + y \sin \frac{5 π}{6} = 2

b

x \cos π + y \sin π = 3

c

x \cos π - y \sin π = 4

d

x \cos \frac{3 π}{2} - y \sin \frac{3 π}{2} = 3

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detailed solution

Correct option is A

$\sqrt{3} x - y + 4 = 0 \Rightarrow - \sqrt{3} x + 1 y = 4$

Divide with 2 an both sides we get

$\frac{- \sqrt{3} x}{2} + \frac{1}{2} y = \frac{4}{2} \Rightarrow x (\frac{- \sqrt{3}}{2}) + y (\frac{1}{2}) = 2$

Here $\cos α = \frac{- \sqrt{3}}{2}, \sin α = \frac{1}{2}, p = 2$

Hence $α$ is in $Q_{2} \Rightarrow α = \frac{5 π}{6}$

$∴$ The equation n of the line in perpendicular form is $x \cos \frac{5 π}{6} + y \sin \frac{5 π}{6} = 2$

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